18.4 Parallel Plate Capacitor
18.4.1 Parallel Plate Capacitor
Fig.1: A commercial capacitor is labeled with the value of its capacitance.
A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance (Fig. 2). We first take the intervening medium between the plates to be vacuum. The effect of a dielectric medium between the plates is discussed in the next section. Let A be the area of each plate and d the separation between them. The two plates have charges Q and –Q. Since d is much smaller than the linear dimension of the plates (d² << A), we can use the result on electric field by an infinite plane sheet of uniform surface charge density.
Fig.2
Plate 1 has surface charge density σ = Q/A and plate 2 has a surface charge density −σ. The electric field in different regions is:
Outer region I (region above the plate 1),
E = σ/(2ε) − σ/(2ε) = 0 (1)
Outer region II (region below the plate 2),
E = σ/(2ε) − σ/(2ε) = 0 (2)
In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving
E = σ/(2ε) + σ/(2ε) = σ/ε
E = Q/(εA) (3)
The direction of electric field is from the positive to the negative plate.
Thus, the electric field is localised between the two plates and is uniform throughout. For plates with finite area, this will not be true near the outer boundaries of the plates. The field lines bend outward at the edges — an effect called ‘fringing of the field’. By the same token, σ will not be strictly uniform on the entire plate. [E and σ are related by E = σ/ε₀.] However, for d² << A, these effects can be ignored in the regions sufficiently far from the edges, and the field there is given by Eq. (3). Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,
V = Ed = Qd/(ε₀A) (4)
The capacitance C of the parallel plate capacitor is then
C = Q/V = ε₀A/d (5)
which, as expected, depends only on the geometry of the system. For typical values like A = 1 m², d = 1 mm, we get
C = (8.85 × 10−12 C²·N−1·m−2 × 1 m²) / (10−3 m) = 8.85 × 10−9 F
(You can check that if 1 F = 1 C·V−1 = 1 C (N·C−1·m)−1 = 1 C²·N−1·m−1.) This shows that 1F is too big a unit in practice, as remarked earlier.
Another way of seeing the ‘bigness’ of 1 F is to calculate the area of the plates needed to have C = 1 F for a separation of, say 1 cm:
A = Cd/ε₀ = (1 F × 10−2 m) / (8.85 × 10−12 C²·N−1·m−2) = 109 m²
which is a plate about 30 km in length and breadth!
Example 18.6
The plates of a parallel-plate capacitor in vacuum are apart and in area 5.00 mm apart and 2.00 m2 in area. A 10.0 kV potential difference is applied across the capacitor. Compute (a) the capacitance; (b) the charge on each plate; and (c) the magnitude of the electric field between the plates.
Answer
We are given the plate area A, the plate spacing d and the potential difference V = 1.00 × 104 V for this parallel-plate capacitor. Our target variables are the capacitance C, the charge on each plate, and the charge Q on each plate, and electric-field magnitude E. We use Eq. (4) to calculate C and then use Eq. (3) and V to find Q. We use E = Q/ε0A to find E.
(a) From Eq.(1)
C = (ε0A)/d
C = (8.85 × 10-12 × 2.00) / (5.00 × 10-3)
C = 3.54 × 10-9 F = 3.54 nF
(b) The charge on the capacitor is
Q = CV
Q = (3.54 × 10-9)(1.00 × 104)
Q = 3.54 × 10-5 C = 35.4 μC
The plate at higher potential has charge +35.4 μC, and the other plate has charge -35.4 μC.
(c) The electric-field magnitude is
E = Q / (ε0A)
E = (3.54 × 10-5) / (8.85 × 10-12 × 2.00)
E = 2.00 × 106 N/C
We can also find E by recalling that the electric field is equal in magnitude to the potential gradient. The field between the plates is uniform, so
E = V/d = (1.00 × 104) / (5.00 × 10-3) = 2.00 × 106 V/m
Example 18.7
A parallel plate capacitor of two metal plates separated by a distance of 5.0 mm. Each plate is a rectangle of size 5.0 cm × 10 cm. The space in between the plates is completely filled with a kind of material. A constant voltage of 12 V is applied across the capacitor. Determine
- (a) the capacitance of the capacitor
- (b) the charge stored by the fully charged capacitor
(relative permittivity of the material = 10; permittivity of free space = 8.85 × 10-12 F m-1)
Answer
(a)
C = (εrε0A)/d
C = (10 × 8.8 × 10-12 × (0.050 × 0.10)) / (5.0 × 10-3)
C = 89 × 10-12 F = 89 pF
(b)
Q = CV
Q = (89 × 10-12)(12)
Q = 1.1 × 10-9 C = 1.1 nC
Example 18.8
Refer to Example 18.7. The material in between the parallel plates is removed so that the space is now filled with air. What is the charge stored by the capacitor?
Answer
Cair = (ε0A)/d
Cair = (1/εr)C
Cair = (1/10)(89 pF) = 8.9 pF
Qair = CairV
Qair = (8.9 pF)(12) = 0.11 nC
Applications of a Parallel Plate Capacitor
- Energy Storage
A parallel plate capacitor is used to store electrical energy in the form of an electric field between its plates. It is commonly used in power supplies and electronic circuits. - Smoothing in Power Supplies
Capacitors are used to smooth out fluctuations in voltage in power supply circuits. They reduce ripples and provide a more stable DC output. - Tuning Circuits
Parallel plate capacitors are used in tuning circuits, such as in radios and televisions, to select specific frequencies. - Touchscreen Technology
They are used in capacitive touchscreens, where the change in capacitance is detected when a finger touches the screen. - Sensors
Used in capacitive sensors to detect changes in distance, pressure, or position by measuring changes in capacitance. - Timing Circuits
Capacitors are used in timing circuits (like RC circuits) to control the timing of signals in electronic devices. - Flash Cameras and Pulsed Power Systems
They store energy and release it quickly, for example in camera flashes or pulsed power applications. - Memory Devices
Used in dynamic random-access memory (DRAM), where capacitors store binary information as electrical charge.
Conclusion: Parallel Plate Capacitor
Parallel plate capacitors are fundamental components in physics and electronics, widely used for storing electrical energy. Their behavior is governed by simple relationships between capacitance, plate area, distance between plates, and the properties of the dielectric material.
The formula C = εA/d clearly shows how capacitance can be controlled and optimized for different applications. By adjusting these parameters, engineers can design capacitors suitable for various purposes.
In practical applications, parallel plate capacitors play an important role in power supplies, electronic circuits, sensors, and modern technologies such as touchscreens and memory devices. Their ability to store and release energy efficiently makes them essential in both basic and advanced electronic systems.
Overall, understanding parallel plate capacitors provides a strong foundation for studying more complex electrical and electronic concepts.
Frequently Asked Questions (FAQ) – Parallel Plate Capacitor
1. What is a parallel plate capacitor?
A parallel plate capacitor is a device consisting of two conductive plates separated by an insulating material (dielectric), used to store electrical energy in an electric field.
2. What is the formula of a parallel plate capacitor?
The capacitance is given by:
C = εA/d
3. How does distance between plates affect capacitance?
Capacitance is inversely proportional to the distance between the plates. Increasing the distance decreases the capacitance.
4. What happens if the plate area increases?
Capacitance increases when the plate area increases because more charge can be stored.
5. What is the role of a dielectric material?
A dielectric increases the capacitance by reducing the electric field strength and allowing more charge to be stored.
6. What is the electric field between the plates?
The electric field between parallel plates is uniform and given by:
E = V/d
7. Where are parallel plate capacitors used?
They are used in power supplies, tuning circuits, sensors, touchscreens, memory devices, and energy storage systems.
8. Can a capacitor store charge permanently?
No, a capacitor cannot store charge permanently. The stored charge will gradually leak away over time.
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